Find the value of the expression 1 + i 2 + i 4 + i 6 + i 8 + … … . + i 20 .

Let y = 1 + i 2 + i 4 + i 6 + i 8 + … … . + i 20 , which is a G . P . with first term 1 and common ratio i 2 . Since, i 2 = - 1 ⇒ y = 1 i 2 11 - 1 i 2 - 1 = - 1 11 - 1 - 1 - 1 ⇒ y = - 1 - 1 - 1 - 1 = 1 .

Express 1 3 + 7 i + ( 9 - 2 i ) - - 4 3 + 2 i in the form a + i b .

Given the expression is : 1 3 + 7 i + ( 9 - 2 i ) - - 4 3 + 2 i = 1 3 + 7 i + 9 - 2 i + 4 3 - 2 i = 1 3 + 9 + 4 3 + i 7 - 2 - 2 = 5 3 + 9 + i 3 = 32 3 + 3 i = a + i b , where a = 32 3 and b = 3 .

Find the conjugate of 1 2 - 7 i .

Let a + i b = 1 2 - 7 i = 2 + 7 i 2 - 7 i 2 + 7 i = 2 + 7 i 2 2 - 7 i 2 = 2 + 7 i 4 - 49 i 2 = 2 + 7 i 4 - 49 - 1 = 2 + 7 i 53 So, the conjugate of a + i b is a - i b = 2 - 7 i 53 .

Find the principal argument of 2 3 - 2 i .

Let z = 2 3 - 2 i Let 2 3 = r cos θ and - 2 = r sin θ By squaring and adding, we get ( 2 3 ) 2 + ( - 2 ) 2 = ( r cos θ ) 2 + ( r sin θ ) 2 ⇒ 12 + 4 = r 2 cos 2 θ + sin 2 θ ⇒ 16 = r 2 ⇒ r = 4 ∴ cos θ = 3 2 and sin θ = - 1 2 ⇒ tan θ = - 1 3 Since, θ lies in fourth quadrant, we have principal argument = - - π 6 = π 6 . Since, θ ∈ ( - π , π ] it is principal argument.

Find the value of ( - 3 ) ( - 6 ) .

Since, i = - 1 ⇒ - 3 - 6 = - 1 3 - 1 6 = i 2 3 6 = - 1 × 18 = - 3 2

Evaluate ( - 2 + 2 - 3 ) 1 2 + ( - 2 - 2 - 3 ) 1 2 .

We have, ( - 2 + 2 - 3 ) 1 2 + ( - 2 - 2 - 3 ) 1 2 = - 2 + 2 3 i 1 2 + - 2 - 2 3 i 1 2 = 1 + 3 i 2 1 2 + 1 - 3 i 2 1 2 = 1 + 3 i + 1 - 3 i = 2 .

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Gamma Question Bank for Engineering Mathematics>Class XI>Chapter 5 - Complex Numbers>Assignment

Embibe Experts Solutions for Exercise 2: Assignment

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 2: Assignment

Attempt the free practice questions from Exercise 2: Assignment with hints and solutions to strengthen your understanding. Gamma Question Bank for Engineering Mathematics solutions are prepared by Experienced Embibe Experts.